In that 2nd equation you wrote, shouldn't it be $Es1=\frac{1}{2}m(v+V)^2$? I mean, only a few words before that you stated that it travels at $v+V$ with respect to s1.
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Alan RomingerOct 17 '11 at 18:29

shouldn't the next to last equation be $\frac{1}{2}m(v+V+dv)^2-\frac{1}{2}m(v+V)^2$?
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Alan RomingerOct 17 '11 at 18:44

2 Answers
2

Let me go ahead and answer the question. I think I have enough information now. We've calculated two expressions for the change in energy, specific to the reference frame. I'm going to use the assumption $dv\ll v$ and $dv \ll V$, because I'm just allergic to heavy algebra like that.

So there we go, yes! The energy change is different depending on the reference frame. But what reference frame was the force exerted from? Imagine that a spaceship with near infinite mass exerted the force to speed up the object and was in the s1 reference frame. I will denote the kinetic energy of the spaceship as $E'$. The change in kinetic energy of the spaceship according to s0 is nothing since with infinite mass the spaceship velocity changes virtually none, and started at nothing.

So there you go. The reason that s0 and s1 disagreed about the change in energy in the object of mass $m$ was because the change in kinetic energy of whatever pushed that object was neglected. Both reference frames agree that the total change in kinetic energy of all objects combined, or $\Delta E+\Delta E'$ is equal to $m v dv$, and this is the amount of energy the spaceship had to spend in order to deliver that impulse to the object with mass $m$. That quantity would change given a different speed of the spaceship.

Here's a knee-jerk answer (I probably should think more before committing to paper): The work done to increase the speed is greater measured in s0 compared to what it is as measured in s1. Whatever force is acting on the particle acts over a longer distance as measured in s0 (but over the same duration of time) than in s1.